91Ó°ÊÓ

We start by rewriting the exponential function \(e^{-\sqrt{x}}\) as a power series. Recall that the power series expansion for \(e^{u}\) is given by: \[e^{u} = \sum_{n=0}^{\infty} \frac{u^n}{n!}\] In our case, we have \(u = -\sqrt{x}\), so we obtain: \[e^{-\sqrt{x}} = \sum_{n=0}^{\infty} \frac{(-\sqrt{x})^n}{n!}\]

Now, we need to integrate the power series term-by-term with respect to the variable x. Given the power series expansion of the given function \(e^{-\sqrt{x}}\), the indefinite integral can be found by integrating each term: \[\int e^{-\sqrt{x}}\, dx = \int \sum_{n=0}^{\infty} \frac{(-\sqrt{x})^n}{n!}\, dx\] We can integrate the series term-by-term: \[\Rightarrow \sum_{n=0}^{\infty} \int \frac{(-\sqrt{x})^n}{n!}\, dx\] Now, let's integrate the term inside the summation: \[\int \frac{(-\sqrt{x})^n}{n!}\, dx = \frac{1}{n!} \int x^{\frac{n}{2}} (-1)^n\, dx\] Using the power rule for integration, this integral becomes: \[\frac{1}{n!} \frac{x^{\frac{n}{2} + 1} (-1)^n}{\frac{n}{2} + 1} + C\] Now substituting this result back into the summation, we get the final answer: \[\int e^{-\sqrt{x}}\, dx = \sum_{n=0}^{\infty} \frac{x^{\frac{n}{2} + 1} (-1)^n}{n!(\frac{n}{2} + 1)} + C\] This is the power series representation for the given indefinite integral.